SB  Efl  2M5 


Q  B 

362 

H5 

1913 

MAIN 


GIFT   OF 


THE 


CAUSE 


OF 


PLANETARY  ROTATION 


ALSO 


A  THEORY 


AS   TO 


THE  TAIL  OF  THE 
COMET 


By  I.  T.  HINTON 


SAN    FRANCISCO 
THE    BLAIR-MURDOCK   COMPANY 


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PLANETARY  ROTATION 


The  following  is  an  attempt  to  assign  the  cause  of  the 
moon  always  presenting  the  same  face  to  the  earth. 

Take  two  small,  equal  masses,  as  indicated  in  the  figure,  at 
equal  distances  from  the  line  of  attraction.  B  attracts  or  is 

attracted  by  a  force  equal  tc__        —,  and  Bf  by  an  equal  force. 

The  components  y  and  • — y  are  equal  and  opposite,  their  alge- 
braic sum  being  equal  to  zero.  The  combined  attraction  of  the 
two  small  masses  in  the  direction  OA  is  equal  to — 


a 


a2  +  3'2        ( a2  +  ?«)*  " "  (a2  -f  y*)»H 
As  either  a  or  v  increases,  the  attraction  will  diminish. 


To  find  the  attractive  force  of  a  circular  disc,  a  resort 
must  be  had  to  calculus.  Let  BBf  be  the  projection  of  the 
disc,  and  suppose  y  to  be  increased  by  a  small  quantity  dy. 

Then  the  differential  of  the  attractive  force  of  the  disc  will  be 
j-rr  dy.    The  integral  of  this  is       2     "  >2 .  ^  ,  and  tak- 

1 


282811 


ing  this  between  the  limits  zero  and  v,  we  obtain  the  attraction 
of  the  circular  disc,  which  is 


This  may  be  written 

<?7rQ2  +  3'2)/2  —  a 
(#+?)* 

Multiplying  both  terms  by  (a2  -\-  v2)^  +  a,  we  obtain 
,?7rj'2  Area  of  disc 


O2  +  r)  +  a  (a-  +  /) 54         ^  (a2  +  r2)  +  j^a  (a2  +  £)* 

This  diminishes  (v  remaining  the  same)  as  a;  increases;  there- 
jfore,  if  we  had  another  disc  on  the  other  side  of  the  center  of 
the  sphere  and  at  the  same  distance  from  that  center,  the 
attraction  of  the  off  disc  would  be  less  than  that  of  the  near 
disc.  Now,  a  sphere  may  be  regarded  as  made  up  of  an 
infinite  number  of  such  discs.  Hence,  the  near  hemisphere 
attracts  with  a  greater  force  than  the  off  hemisphere. 

A  more  satisfactory  proof  of  this,  perhaps,  is  the  following : 
When  the  disc  is  taken  as  a  section  of  a  homogeneous  sphere, 
and  c  is  the  distance  apart  of  the  centers,  the  distance  a  be- 
comes c  +  .r  and  v2  becomes  r2  —  .r2.  Substituting  these  values 
in  {A)  we  have 


(2- 
V 


If  this  be  multiplied  by  dx,  and  the  integral  be  taken  between 
the  limits  —  r  and  -J-  r,  we  shall  have  the  attraction  of  the 
entire  sphere.  Dropping  TT  for  the  present  to  simplify  matters, 
we  have 

,  scdx  2x$x 

(c*  +  2cx  +  r*)X          (cz  +  sex '+ r*)* 

Integrating  (using  the  formula  in<  =  (ndi*  -j-  (i'dii  for  the 
last  term)  we  have 

(c-  -f  sex  -4-  r-y/* 

2X  —  2(C~  +  2CX  +  P*)*  —  2X  - 


Making 
x  =  +  r 


Subtract- 
ing 


+  2r  — 


-4^)+ 
w^-.u 


4T—4T 


-4r  + 


3? 


3c' 


(c  +  r)3  =  c3  +  jrV  +  3cr-  +  r3 
( c  —  n.) 3  =  c3  —  jc2r  +  jcr2  —  r3 


That  is  to  say,  a  homogeneous  sphere  attracts  or  is  attracted 
as  if  its  entire  mass  were  concentrated  at  its  center.  (The 
geometrical  demonstration  of  this  truth  given  in  the  books 
is  faulty.)  But  it  is  not  a  fact  that  the  two  hemispheres  are 
attracted  equally.  To  prove  this,  let  us  take  the  integral  first 
from  —  r  to  o,  and  then  from  o  to  +  r.  The  integral  (B)  is : 


2X  -  2  (  C-  -  2CX  -\-r2)1/2 


2X 


(c2  + 


Making 
x  =  O 


.v  =  —  r 


Subtract- 
ing 


2Y 2(C r)   -f  2Y 


(c  —  r) 


(V2    l    rz\iy2 
±     ^  8  ^ 


2 


Which,  multiplied  by  TT,  is  the  attraction  of  the  near  hemisphere. 

3 


Making 


.1-  =  0 


Subtract- 
ing 


2Y  —  2(c  —  r)  — 


0  —  ^    c2 


-2c  +  2(e- 


^     +2^- 


2(e- 


[u  + 


(c  +  r) 


•2-f  r=)'5*]     .    (b) 


For  the  attraction  of  the  off  hemisphere.    Adding  (a)  and  (b), 


we  obtain 


,  as  a  check.     To  prove  algebraically  that  (a) 


is  greater  than  (b)  would  lead  to  complexity;  hence  a  resort 
will  be  had  to  arithmetic,  making  c  =  12  and  r  =  5.  Sub- 
stituting these  values  in  (a),  the  result  will  be: 


12 


43 


24 


—  26 


70 

12 


-h  -750 


12  )  70  (  5.833 
60 

IOO 

96 

40 


13 

507 
169 

133  =  2197 


73  =  343 


=  2197 
=    343 

216  )  1854  (  8.583 
1728 

1260 
1080 


1800 
1728 

720 


And  the  attraction  of  the  near  hemisphere  is  .750. 

4 


For  the  off  hemisphere  the  result  will  be : 
-  2c  +  2(c*  +  r*)*  —  2r—       -+-^-[  (c  +  r)8  — C^  +  r2)1*]    (b) 


12  )  170  (  I4.i6?  17 

12  17 


50  119 

48  17 


20  289 

12  17 


80  2023 

72  289 

__,  OA  

+  26  i;3  : 

—  I4.I67 
+  12-574 


13 
13 

4913 
2197 

216  )  2716  (  12.574 
216 

39 
13 

169 
13 

556 
432 

507 
169 

1240 
1080 

I33  =  2197 

1600 
1512 

-f  .407 


And  the  attraction  of  the  off  hemisphere  is  -[-.407. 

4r3  ___  500 

jr2  "  ~  432  Attraction  of  whole  sphere. 

432  )  500  (  1.157  Near  hemisphere  ..............  750 

432  Off  hemisphere  ...............  407 

680 

432  Whole  sphere  ............    1.157 

2480 
2160 


3200 
3024 


Now,  suppose  a  body  to  be  at  the  point  C,  and  moving  at 
right  angles  to  OC,  so  that  it  will  travel  to  Cl  in  a  certain  in- 
terval of  time,  as  indicated  in  the  figure,  C  being  the  center 
of  the  sphere,  and  A  and  B  being  the  centers  of  the  mass  of 
the  respective  hemispheres.  If  there  were  no  force  other  than 
this  tangential  force,  the  center  C  would  move  to  C\.  Suppose, 


again,  that  afterwards  an  attractive  force  acts  for  the  same 
length  of  time  in  the  direction  OCl.  If  we  regard  the  forces 
as  acting  on  the  whole  sphere,  the  center  will  take  some  such 
position  as  C., ;  and  if  we  regard  the  forces  as  acting  upon 
the  two  hemispheres  separately,  the  points  A  and  B  will  assume 


the  positions  A,,  and  B2.  The  component  distances  AA2, 
CjC2,  B±B2,  and  A2A3,  C2C3,  B2B3  are  not  in  the  above  ratio 
•75°  :  /^  (1.157)  '•  -4°7>  though  nearly  so.  However,  the  smal- 
ler the  intervals  of  time  and  space  are  taken,  the  more  nearly 
Will  the  ratios  approach  equality,  until,  when  the  calculus  limit 
is  reached  and  the  orbit  becomes  a  curve,  instead  of  a  polygon, 
the  lines  OA,  OC,  OB  will  coincide  within  an  infinitely  small 
angle,  and  the  lines  A3A2,  C3C2,  B3B2  will  attain  the  ratio 
.750:^2(1.157)  1.407.  In  other  words,  the  original  points 
A,  C,  B  will  assume  the  position  OA2C2B2,  all  on  the  same 
straight  line.  This  clearly  indicates  a  rotation  going  part  />a^w 
with  the  revolution.  Hence,  when  gravity  acts  as  a  central 
acceleration,  the  secondary  will  always  present  the  same  face 
to  the  primary.  Q.  E.  D. 


The  earth  does  not  rotate  according  to  this  law,  but  here 
other  forces  have  been  or  are  now  in  action.  The  following  is 
a  highly  improbable,  but  may  be  possible  explanation:  If  the 
earth  had  as  regards  the  sun  no  rotation  at  all,  or  a  rotation 
(due  to  the  moon's  influence)  once  in  about  twenty-eight  days, 
the  face  towards  the  sun  would  soon  become  very  hot,  and 


the  water  on  that  side  would  evaporate  rapidly  and  condense 
as  ice  on  the  surface  of  the  earth's  off  hemisphere.  After  a 
while  the  condition  would  be  somewhat  as  in  the  figure,  the 
shaded  portion  representing  ice. 

Such  a  body  would  (I  think)  rotate  with  a  continuous  ac- 
celeration, the  ratio  of  acceleration  diminishing  as  the  ice 
melted  faster  through  the  more  frequent  turning  of  the  earth's 
face  to  the  sun,  until  the  acceleration  became  nil.  The  body 
would  then  continue  to  rotate  at  the  speed  attained  when  the 
acceleration  became  zero. 

WEIGHING   THE    EARTH. 

If  a  known  weight  were  in  a  well  at  the  point  A  in  the  figure 
on  page  I,  its  attraction  towards  the  center  would  be  equal 
to  the  attraction  of  the  mass  BEB'  less  that  of  the  mass  BDB' ' , 
which  can  be  found  by  integration,  and  the  attraction  or 
weight  of  the  earth  could  be  found  if  we  knew  the  weight 
of  the  spherical  segment  BDB' .  This  would  be  exceedingly 
difficult  to  obtain  with  any  degree  of  accuracy  on  the  land,  but 
at  sea  we  can  go  down  five  miles,  and  the  weight  of  a  spherical 
segment  of  water  whose  height  is  five  miles  can  be  calculated 
to  a  nicety. 


THE   TAIL   OF   THE   COMET. 

In  endeavoring  to  explain  this  phenomenon  by  the  laws  of 
mechanics,  the  writer  chanced  upon  the  preceding  demon- 
stration, without  making  any  progress  on  the  main  problem. 
However,  he  ventures  the  following  explanation  of  the  comet's 
tail  from  physical  laws: 

The  nucleus  of  the  comet,  it  appears  to  be  admitted,  is  a 
mass  of  very  hot  vapor  or  vapors,  and  if  this  is  true  it  must 
be  surrounded  by  a  vast  mass  of  less  hot  and  less  condensed 
vapors,  this  large  mass,  or  a  portion  thereof  vastly  larger 
than  the  nucleus,  being  somtimes  faintly  visible  by  starlight. 
The  comet  is  therefore  an  illuminated  nucleus  surrounded  by 
an  immense  envelope  of  vapor,  hot,  but  not  as  hot  as  the 
nucleus.  If  a  cannon-ball  were  lying  on  the  surface  of  the 
earth,  it  would  weigh  more  at  12  midnight  than  at  noon,  since 
in  the  one  case  the  sun's  attraction  would  be  added  to  gravity 
and  in  the  other  it  would  be  subtracted  from  it.  The  moon 
at  conjunction  would  have  a  greater  effect,  perhaps  enough 
to  be  noticed  on  a  fine  spring  balance.  So,  the  comet  would 
be  flattened  on  the  side  away  from  the  sun,  and  elongated  on 
the  near  side,  thus  assuming  an  egg-like  form.  However, 
when  the  distance  is  great,  the  form  is  probably  nearly  a 
sphere.  Now,  the  nucleus  of  a  comet  is  dense  enough  to  cast 
a  shadow,  and  against  the  dark  background  of  that  shadow  the 
heated  particles  immediately  surrounding  that  shadow  be- 
come visible,  while  all  the  rest  of  the  vast  volume  of  the  comet 
except  the  nucleus  fail  to  give  out  sufficient  light  to  be  visible. 
Their  light  is  obscured  just  as  that  of  the  moon  is  obscured  in 
the  daytime,  while  the  particles  around  the  shadow  of  the 
nucleus  are  more  bright  from  the  same  reason  that  the  corona 
of  the  sun  appears  brighter  during  an  eclipse.  The  length  of 
the  tail  depends  on  the  heat  of  the  comet,  the  distance  from 
the  sun  and  the  angle  of  view  from  the  earth.  The  curvature 
is  due  to  the  difference  in  time  taken  by  the  light  coming  from 
the  nucleus  and  from  the  tail.  It  is  most  noticeable  when  the 
comet  is  near  the  sun,  since  then  the  velocity  of  the  nucleus  is 
immense,  and  that  of  the  end  of  its  shadow  is  very  much 
more  so. 

8 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


i  ^  i  ar 

/ILI\ 

ICL^ 

IN) 

iYtrrisi  1887 

~ 

/  * 

RECEIVED 

DEC  ^    '67  -P  M 

LOAN  PEPT 

LD  2lA-60m-2.'67 
(H241slO)476B 


General  Library 

University  of  California 

Berkeley 


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